Weak convergence in $L^2(0,T;L^2)$ and a $\limsup$

Question

Let $u_n \rightharpoonup u$ in $L^2(0,T;L^2(\Omega))$ for a bounded domain $\Omega$.

We have $$\lVert u_n(t) \rVert_{L^2(\Omega)} \leq C$$ for all $t$ where $C$ is independent of $t$.

Does it follow that $$\limsup_n\int_0^T w\lVert u_n(t) \rVert_{L^2(\Omega)}\;dt \geq \int_0^T w\lVert u(t) \rVert_{L^2(\Omega)}\;dt$$ given $w \in L^2(0,T)$?

I have no clue how to proceed. Unfortunately the bound on $u_n(t)$ only gives $u_{n_j(t)}(t)$ weak convergence to $u(t)$ (the subsequence depends on $t$).

EDIT: If it helps, we have $u_n \to u$ in $L^2(0,T;H^{-1}(\Omega))$ (yes I do mean $H^{-1}$).

Answer 1

This holds iff $w \ge 0$ a.e.

In this case, the functional $u \mapsto \int_0^T w(t) \, \|u(t)\|_{L^2(\Omega)} \, \mathrm{d}t$ is convex and continuous, hence weakly lower semicontinuous.

If $w < 0$ a.e. on a subset $I$ of positive measure, take your favorite sequence $\{v_n\} \subset L^2(\Omega)$ with $v_n \rightharpoonup 0$, but $v_n \not\to 0$ in $L^2(\Omega)$. Then take $u_n(t,x) = \chi_I(t) \, v_n(x)$.